Entropy Lesson Notes PDF

LEARNING TARGETS ▪ I can explain how changes in energy and changes in entropy influence the spontaneity of a chemical reaction ▪ I can analyze the entropy change of a chemical reactions using the standard value of entropy Entropy SECOND LAW OF THERMODYNAMICS - spontaneous change is accomp...

LEARNING TARGETS ▪ I can explain how changes in energy and changes in entropy influence the spontaneity of a chemical reaction ▪ I can analyze the entropy change of a chemical reactions using the standard value of entropy Entropy SECOND LAW OF THERMODYNAMICS - spontaneous change is accompanied by increase in entropy while other spontaneous systems occur with a decrease in entropy - if both system and the surroundings are considered, “all naturally occurring processes occurs spontaneously toward the directions of the process that increases entropy of the universe Entropy SECOND LAW OF THERMODYNAMICS ∆Suniverse = [ ∆Ssystem + ∆Ssurrounding ] ≥ 0 ✓ the ∆S universe determined from the sum of ∆S system and ∆S surrounding must be a positive value which corresponds to an increase in entropy ✓ a process is considered spontaneous only if the total change in entropy of the universe is an increase in entropy ENTROPY CHANGE of a REACTION/ SYSTEM The exact entropy change for the process can be determined in the same way as the enthalpy change for the reaction using standard absolute entropies instead of enthalpies of formation. For any chemical reaction, ∆Soreaction/system = ∑Soproducts - ∑Soreactant ENTROPY CHANGE of SURROUNDING The entropy change in the surroundings is directly proportional to the enthalpy change of the reaction and inversely proportional to the absolute temperature (in Kelvin) of the surroundings. −∆Hrxn ∆Sosurrounding = T ENTROPY Sample Problem Calculate the change in entropy at standard state and 25oC that accompanies the reaction C(s, graphite) + CO2(g) → 2 CO(g) Absolute molar entropies at 25oC C(s, graphite) = 5.740 J/mol-K CO2(g) = 213.74 J/mol-K CO(g) = 197.67 J/mol-K ENTROPY ∆Soreaction = ∑Soproducts - ∑Soreactant = (2 moles x 197.67 J/mol-K) – [(1mole x 5.740 J/mol-K) + (1 mole x 213.74 J/mol-K)] = 395.34 J/K – 219.48 J/K = 175.86 J/K ENTROPY ∆Horxn = [ ∑Hf(products) - ∑Hf(reactants) ] = [(2)(-110.5 kJ/mol) – (-393.5 kJ/mol + 0)] ∆Horxn = 172.5 kJ −∆Hrxn ∆Sosurrounding = T − (172500 J) = = -578.86 J/K 298 K ENTROPY To get the total entropy of the process (∆Suniverse) ∆Suniverse = [ ∆Ssystem + ∆Ssurrounding ] = [175.86 J/K + (-578.86 J/K) ] ∆Suniverse = - 403 J/K Decrease in the total entropy, non-spontaneous process ENTROPY Calculate the total entropy, ∆Suniverse for each given reaction. Tell whether the process is spontaneous or not at 25oC. 1. 2H2(g) + O2(g) → 2H2O(g) Substance H2(g) O2(g) H2O(g) So(J/mol-K) 130.6 205.0 188.7 Hfo (kJ/mol) 0 0 -241.8 ENTROPY ∆Soreaction = ∑Soproducts - ∑Soreactant = (2 moles x 188.7 J/mol-K) – [(2 moles x 130.6 J/mol-K) + (1 mole x 205.0 J/mol-K)] = 377.4 J/K – 466.2 J/K ∆Soreaction = -88.8 J/K ENTROPY ∆Horxn = [ ∑Hf(products) - ∑Hf(reactants) ] = [(2)(-241.8 kJ/mol) – (0 + 0)] ∆Horxn = - 483.6 kJ −∆Hrxn ∆Sosurrounding = T − (−483600 J) = = 1 622.82 J/K 298 K ENTROPY To get the total entropy of the process (∆Suniverse) ∆Suniverse = [ ∆Ssystem + ∆Ssurrounding ] = [-88.8 J/K + 1 622.82 J/K] ∆Suniverse = 1 534.02 J/K Increase in the total entropy, spontaneous process Formative Assessment Calculate the total entropy, ∆Suniverse for each given reaction. Tell whether the process is spontaneous or not at 25oC. 2. N2(g) + 3H2(g) → 2NH3(g) Substance N2(g) H2(g) NH3(g) So(J/mol-K) 191.5 130.6 192.3 Hfo (kJ/mol) 0 0 -46.3 Activity Calculate the total entropy, ∆Suniverse for each given reaction. Tell whether the process is spontaneous or not at 25oC. 2KClO3(s) → 2KCl(s) + 3O2(g) Substance KClO3(s) KCl(s) O2(g) So(J/mol-K) 143.1 82.6 205.0 Hfo (kJ/mol) -397.7 -436.77 0